To be straight , this is a home - work question .

This is the question.

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Okay, now let me give you the my attempt and what I believe to be correct.

  • Here E.M.F induced across OA is constant, because B field is spreaded throughout the conductor, or mathematically I can argue as below

$$\Delta \Phi = B \cdot dA = B \frac{\omega\; \Delta t}{2\pi} \pi r^2$$

$$\frac{d\Phi}{dt} = \frac{B\omega r^2}{2}$$

Therefore I can say induced E.M.F is constant.

So the answer should be one of (1) and (4).

  • But how to find the induced current, I am totally stuck on it. But I have one clue that induced current is indirectly proportional to resistance. Because induced emf is constant.But how to proceed? I can not see a clear path.

The emf is constant. You know that current is voltage divided by resistance. You also know that if the current changes, it changes the flux through the loop (the more current flows the more you reduce the B field inside the loop)

As you go around the loop, what happens to the resistance of the circuit?

Once you have figured that out you can answer the question yourself.

  • \$\begingroup\$ is the resistance remain constant ? \$\endgroup\$ – On the way to success Jul 6 '15 at 5:25

You are completely right the emf will remain constant as the speed the arm is moving around is constant.

Therefore you only need to look at the resistance of the circuit in order to be able to determine the current.

Try thinking about the circuit as a basic potentiometer and think about the speed of the arm is moving around to work out how the resistance will change as the arm moves round. Then simply translate this into the current using Ohms law.

  • \$\begingroup\$ Do I need to consider resistance of the arc PQR or the OA? \$\endgroup\$ – On the way to success Jul 6 '15 at 11:20
  • \$\begingroup\$ Well both are part of the circuit so both contribute to the resistance, however the length of the arc that is part of the circuit is what you should be concerned with in terms of looking at how it varies \$\endgroup\$ – Calum Jul 6 '15 at 11:36

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